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Root of (16*20+8*32)=

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Bunuel
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Root of (16*20+8*32)= [#permalink] New post   03 Sep 2012, 06:00
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00:00
A
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15% (low)

Question Stats:

81% (01:29) correct 19% (02:18) wrong based on 3334 sessions
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\(\sqrt{16*20+8*32}=\)

(A) \(4\sqrt{20}\)
(B) 24
(C) 25
(D) \(4\sqrt{20}+8\sqrt{2}\)
(E) 32
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B
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Bunuel
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Re: Root of (16*20+8*32)= [#permalink] New post   03 Sep 2012, 06:01
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SOLUTION

\(\sqrt{16*20+8*32}=\)

(A) \(4\sqrt{20}\)
(B) 24
(C) 25
(D) \(4\sqrt{20}+8\sqrt{2}\)
(E) 32

Factor out 16: \(\sqrt{16*20+8*32}=\sqrt{16(20+8*2)}=\sqrt{16*36}=4*6=24\).

Answer: B.
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Re: Root of (16*20+8*32)= [#permalink] New post   15 Oct 2012, 03:02
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asveaass wrote:
Obviously I don't get this..... :shock:

I know this must be wrong since no answer choice allows this, but where am I mistaking? Appreciate the help!

Rt (16*20)+(8*32)=
Rt(2^4*2^2*5)+(2^3*2^5)=
2^3 Rt(5)+2^4=
8Rt5+16 ??


It should be:

\(\sqrt{16*20+8*32}=\sqrt{2^4*2^2*5+2^3*2^5}=\sqrt{2^6*5+2^8}=\sqrt{2^6(5+2^2)}=\sqrt{2^6*9}=2^3*3=24\).

Hope it's clear.
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Re: Root of (16*20+8*32)= [#permalink] New post   03 Sep 2012, 07:56
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\sqrt{4*4*4*5 + 4*4*4*4} = 2*2*2\sqrt{5+4} = 8*\sqrt{9} = 8*3 = 24

B
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Re: Root of (16*20+8*32)= [#permalink] New post   14 Oct 2012, 08:05
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Obviously I don't get this..... :shock:

I know this must be wrong since no answer choice allows this, but where am I mistaking? Appreciate the help!

Rt (16*20)+(8*32)=
Rt(2^4*2^2*5)+(2^3*2^5)=
2^3 Rt(5)+2^4=
8Rt5+16 ??
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Re: Root of (16*20+8*32)= [#permalink] New post   28 Feb 2015, 18:15
why can't you factor out a 4? it doesnt work when I do it.
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Re: Root of (16*20+8*32)= [#permalink] New post   28 Feb 2015, 18:52
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Hi soniasawhney,

You CAN factor out a 4, but I'd like to see your work so that we can assess whether you're factoring out that 4 correctly or not. So, can you post your work/"steps"?

As an aside, after factoring out a 4, you'd then have more work to do to get to the solution.

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Re: Root of (16*20+8*32)= [#permalink] New post   01 Mar 2015, 13:27
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This is how I was doing it..

4[(4*5)+(2*8)]
4[20+16]
4[36]
take the square root...
(2)(6)=12

However, I now realize that the way I was factoring it, I actually was taking out 'too many' 4's and I should have done it like so...is this right?
4[(16*5)+(8*8)]
4[80+64]
4[144]
take the square root...
2*12=24

Let me know if that makes sense!
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Re: Root of (16*20+8*32)= [#permalink] New post   01 Mar 2015, 14:52
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Hi soniasawhney,

As you've come to realize, you have to be careful when factoring out numbers from large calculations.

There are actually several different ways to 'factor down' and simplify this question, but your second approach is correct (and is just as valid as any other).

Here's another way to do it (based on the same ideas that you were using):

You might catch that 16 is a factor of BOTH terms (16x20 and 8x32). By factoring out 16, we can simplify the calculation even further....

Root[(16)(20) + (8)(32)]

Root[16[(20) + (8)(2)]
Root[16[(20) + 16]]
Root[16[36]]

4(6) = 24

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Re: Root of (16*20+8*32)= [#permalink] New post   14 Aug 2015, 20:58
why is 16 only factored from 2 and the 20 and 8 are left alone? sorry if this is a bad question, I'm new to the gmat and need every explanation i could get
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Re: Root of (16*20+8*32)= [#permalink] New post   15 Aug 2015, 10:11
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Hi faizanswx,

When dealing with radicals, you should look for 'perfect squares' (re: 4, 9, 16, 25, etc.) that you can factor-out of the radical...

eg. √50 = √(25x2) = 5√2

In the given prompt, 16 can be factored-out (it becomes '4') and then the 36 can factored-out (it becomes '6').

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Re: Root of (16*20+8*32)= [#permalink] New post   31 Oct 2015, 23:43
Ok, this problem is completely baffling the crap out of me! lol!!

Let me first state an assumption I made when doing this problem:

Step 1: Sqrt [(16*20)+(8*32)]= sqrt(16*20) + sqrt(8*32)

Step 2: Sqrt (16*20) = 4*Sqrt(20) = is atleast 16 & sqrt(8*32) = 16

So how in the world is this answer not atleast 32 (I know 32 is not in the answer choice) ? or did I make a big mistake in step 1?
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Re: Root of (16*20+8*32)= [#permalink] New post   01 Nov 2015, 02:08
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EricImasogie wrote:
Ok, this problem is completely baffling the crap out of me! lol!!

Let me first state an assumption I made when doing this problem:

Step 1: Sqrt [(16*20)+(8*32)]= sqrt(16*20) + sqrt(8*32)

Step 2: Sqrt (16*20) = 4*Sqrt(20) = is atleast 16 & sqrt(8*32) = 16

So how in the world is this answer not atleast 32 (I know 32 is not in the answer choice) ? or did I make a big mistake in step 1?


You should brush-up fundamentals before attempting the questions.

Generally \(\sqrt{a+b}\neq \sqrt{a} + \sqrt{b}\). For example, \((\sqrt{4+4}=\sqrt{8})\neq (\sqrt{4} + \sqrt{4}=4)\).

Theory on Exponents: math-number-theory-88376.html
Tips on Exponents: exponents-and-roots-on-the-gmat-tips-and-hints-174993.html

All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39
All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60

Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html
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Re: Root of (16*20+8*32)= [#permalink] New post   01 Nov 2015, 16:12
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Hi EricImasogie,

While the major Radical rules won't show up too often on Test Day (you'll likely see just 1-2 questions that will involve radical rules), they are a standard part of the GMAT, so if you know the rules, then you should be able to pick up these points.

Since your 'assumptions' about how the math 'worked' were incorrect, I have a few questions about your studies so far:

1) How long have you been studying?
2) What resources have you been using?

It might be that you need some general 'math' practice before you do too much more GMAT practice.

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Re: Root of (16*20+8*32)= [#permalink] New post   27 Mar 2016, 20:05
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Here is a visual that should help.
Attachments

Screen Shot 2016-03-27 at 8.04.32 PM.png
Screen Shot 2016-03-27 at 8.04.32 PM.png [ 105.52 KiB | Viewed 26199 times ]

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Re: Root of (16*20+8*32)= [#permalink] New post   04 May 2016, 09:41
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Bunuel wrote:
\(\sqrt{16*20+8*32}=\)

(A) \(4\sqrt{20}\)
(B) 24
(C) 25
(D) \(4\sqrt{20}+8\sqrt{2}\)
(E) 32



Solution:

We must first simplify the expression in the square root before actually taking the square root. In other words, we have to get the product of 16 and 20 and add it to the product of 8 and 32 before taking the square root.

√[(16)(20) + (8)(32)]

√(320 + 256)

√576 = 24

The answer is B.

Note: If struggling with determining the value of √576, we could have used the answer choices to our advantage. We need to ask ourselves what number, when squared, equals 576. Since we should have the value of 25^2 memorized, we would know that 25^2 = 625. Since 576 is slightly less than 625, we can reasonably determine that 24^2 = 576.
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Re: Root of (16*20+8*32)= [#permalink] New post   Updated on: 09 Jun 2016, 02:34
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The first thing when you see such questions should be to simplify
16*20 + 8*32 = 320 + 256 = 576
Now, \(\sqrt{576}\)= 24

Correct Option: B

Originally posted by OptimusPrepJanielle on 08 Jun 2016, 20:42.
Last edited by OptimusPrepJanielle on 09 Jun 2016, 02:34, edited 1 time in total.
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Re: Root of (16*20+8*32)= [#permalink] New post   11 Oct 2019, 08:53
Bunuel wrote:
\(\sqrt{16*20+8*32}=\)

(A) \(4\sqrt{20}\)
(B) 24
(C) 25
(D) \(4\sqrt{20}+8\sqrt{2}\)
(E) 32

Practice Questions
Question: 35
Page: 156
Difficulty: 600


\(\sqrt{16*20+8*32}\)

= \(\sqrt{320+256}\)

= \(\sqrt{576}\)

\(24 * 24 = 576\) , Hence Answer must be (B)
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Re: Root of (16*20+8*32)= [#permalink] New post   06 Feb 2024, 17:36
Bunuel wrote:
asveaass wrote:
:shock:

I know this must be wrong since no answer choice allows this, but where am I mistaking? Appreciate the help!

Rt (16*20)+(8*32)=
Rt(2^4*2^2*5)+(2^3*2^5)=
2^3 Rt(5)+2^4=
8Rt5+16 ??


It should be:

\(\sqrt{16*20+8*32}=\sqrt{2^4*2^2*5+2^3*2^5}=\sqrt{2^6*5+2^8}=\sqrt{2^6(5+2^2)}=\sqrt{2^6*9}=2^3*3=24\).

Hope it's clear.

­so whenever we have an exponent inside a square root, we didvide by 2? 
eg: square root of 2^8 gives 2^4?
square root of  4^6 gives 4^3?
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Re: Root of (16*20+8*32)= [#permalink] New post   07 Feb 2024, 00:08
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Thib33600 wrote:
Bunuel wrote:
asveaass wrote:
:shock:

I know this must be wrong since no answer choice allows this, but where am I mistaking? Appreciate the help!

Rt (16*20)+(8*32)=
Rt(2^4*2^2*5)+(2^3*2^5)=
2^3 Rt(5)+2^4=
8Rt5+16 ??


It should be:

\(\sqrt{16*20+8*32}=\sqrt{2^4*2^2*5+2^3*2^5}=\sqrt{2^6*5+2^8}=\sqrt{2^6(5+2^2)}=\sqrt{2^6*9}=2^3*3=24\).

Hope it's clear.

­so whenever we have an exponent inside a square root, we didvide by 2? 
eg: square root of 2^8 gives 2^4?
square root of  4^6 gives 4^3?

­
Yes, the square root of a number, \(\sqrt{x}\), means \(x^{\frac{1}{2}}\), and thus essentially says to multiply the exponent of x by 1/2. Similarly, \(\sqrt[3]{x}\) means \(x^{\frac{1}{3}}\), and thus essentially says to multiply the exponent of x by 1/3. And so on.

8. Exponents and Roots of Numbers


Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.­­­­­
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Re: Root of (16*20+8*32)= [#permalink]
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